Planetary Ballistics

The escape velocity is $v_e \; = \; \sqrt{\frac{2GM}{R}}$ where $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet. Since all the particle's motion lies outside the planet's radius, we can assume that the planet's attraction is as from a point particle, and so the equation of motion of the particle is $m\ddot{\mathrm{r}} \; = \; -\frac{GMm}{r^2}\hat{\mathbf{r}} \; = \; -\frac{mRv_e^2}{2r^2}\hat{\mathbf{r}}$ where $m$ is the mass of the projectile. Thus $\ddot{\mathbf{r}} \; = \; -\frac{Rv_e^2}{2r^2}\hat{\mathbf{r}}$

The trajectory of the particle is an ellipse with one focus being the origin. By symmetry, the major axis is along the line $y=x$ . Since the particle's velocity is parallel to the vector ${1 \choose 1}$ at the point of projection, we deduce that the point of projection and the point of collision with the earth are the opposite ends of the minor axis. Thus (in the standard notation) $b = ae = \tfrac{1}{\sqrt{2}}R$ , and hence $e = \tfrac{1}{\sqrt{2}}$ and $a=R$ .

Using plane polar coordinates as shown in the diagram, standard theory shows that $\frac{1}{r} \; = \; \frac{v_e^2}{Rv_0^2}\big(1 + e\cos\theta\big) \; = \;\frac{1}{R\alpha^2}\left(1 + \frac{1}{\sqrt{2}}\cos\theta\right)$ When $\theta = \tfrac34\pi$ we have $r = R$ , and hence $\alpha = \boxed{\tfrac{1}{\sqrt{2}}}$ .